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Alik [6]
1 year ago
6

Please I need help Asap!

Mathematics
1 answer:
Reil [10]1 year ago
6 0

Answer:

-11/12

Step-by-step explanation:

\frac{5}{8}+\frac{3}{4}=\frac{5}{8}+\frac{6}{8}=\frac{11}{8} \\ \\ -\frac{2}{3}-\frac{5}{6}=-\frac{4}{6}-\frac{5}{6}=-\frac{9}{6}=-\frac{3}{2} \\ \\ \frac{11/8}{-3/2}=\frac{22/8}{-3}=-\frac{22}{24}=-\frac{11}{12}

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Answer:

(x + 7)^2 + (y + 3)^2 = 4

Step-by-step explanation:

The standard equation of a circle is:

(x - h)^2 + (y - k)^2 = r^2

where (h, k) is the center and r is the radius. We can substitute the information given to obtain:

(x + 7)^2 + (y + 3)^2 = 4

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3 years ago
What do you do on Brainly?
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Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2
leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where \lambda = \frac{10.2}{200} = 0.051

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

\frac{1}{4} \times 200 = 50

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004

c)The expected number of grams to be eaten before encountering the first fragments :

E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 grams

7 0
3 years ago
Calculate the volume HELP PLEASE ASAP
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r²= 9

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= 0.3 x 4 x 3.14 x 9

= 33.912
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2 years ago
4. A dolphin decsends toward the bottom of
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It descended 240 feet 12*20 is 240
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